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Let $y=y=\sqrt{e^\sqrt x },x>0\;find\;\large\frac{dy}{dx}.$

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  • Use chain rule.
$ \large\frac{dy}{dx}\: \large\frac{1}{2\sqrt{e^{\sqrt x}}}$ x $e^{\sqrt x}$ x $ \large\frac{1}{2\sqrt x}= \large\frac{\sqrt{e^{\sqrt x}}}{4\sqrt x}$

 

answered Mar 12, 2013 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 

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